Question: The number $2^{1993}+3^{1993}$ is a multiple of $5.$ What is the units digit of the quotient $\frac{2^{1993}+3^{1993}}{5}?$
Solution: We want to find the units digit of the quotient \[\frac{2^{1993}+3^{1993}}5.\]We list the final two digits of  $2^n$ and $3^n$ in the next table. We also compute the units digit of the quotient whenever $2^n+3^n$ is divisible by $5.$

\begin{tabular}{|c|c|c|c|c|}
\hline
$n$&$2^n$&$3^n$&$2^n+3^n$&$\frac{2^n+3^n}5$\\
\hline
0&01&01&02&\\
1&02&03&05&1\\
2&04&09&13&\\
3&08&27&35&7\\
4&16&81&97&\\
5&32&43&75&5\\
6&64&29&93&\\
7&28&87&15&3\\
8&56&61&17&\\
9&12&83&95&9\\
10&24&49&73&\\
11&48&47&95&9\\
12&96&41&37&\\
13&92&23&15&3\\
14&84&69&53&\\
15&68&07&75&5\\
16&36&21&57&\\
17&72&63&35&7\\
18&44&89&33&\\
19&88&67&55&1\\
20&76&01&77&\\
21&52&03&55&1\\
22&04&09&13&\\
23&08&27&35&7\\
24&16&81&97&\\
25&32&43&75&5\\
\hline
\end{tabular}We notice that after the first pair, the sequence repeats every $20.$ Therefore \[{2^{1993}+3^{1993}}\equiv {2^{13}+3^{13}}\equiv15\pmod{100}.\]So, the units digit of the quotient $\frac{2^{1993}+3^{1993}}5$ is $\boxed{3}.$

(Note: "mod 100" essentially means "remainder when the number is divided by 100".  So, $2^{1993} + 3^{1993} \equiv 15 \pmod{100}$ means that $2^{1993} + 3^{1993}$ is 15 more than a multiple of 100.)